# Next we show that the sequence is increasing this can

• Notes
• 4

This preview shows pages 3–4. Sign up to view the full content.

Next we show that the sequence is increasing. This can be done again by induction, but it is also possible to do it directly, using the bounds we obtained in the first step. Recall that if y R , 0 y < 1, we have y y . Since we showed 2 x n < 3, we thus have 0 x n - 2 < 1, so x n - 2 x n - 2. This implies that x n +1 = 2 + x n - 2 x n , for an arbitrary n N , thus our sequence is increasing. Note also that if x 1 = 2, then x n = 2 for any n (can be shown immediately by induction), so the sequence is constant, hence trivially convergent to 2. If x 1 > 2, then all inequalities are strict, so our sequence is strictly increasing. We showed that { x n } n is bounded and increasing, so by the monotone convergence theorem, the sequence is convergent. Let’s denote the limit with L . Taking the limit as n → ∞ in the recursive relation, we get L = 2 + L - 2. Solving we get two solutions L 1 = 2 and L 2 = 3. As we noted above, if x 1 = 2, the sequence is constant 2, so the limit is 2. If x 1 > 2, then 2 < x 1 < x 2 < ... < x n < ... < 3, so the limit is the supremum of the sequence, and the supremum cannot be 2. Thus the limit is 3. Observation: Another solution, perhaps even shorter, can be obtained by showing from the recursive relation that x n = 2 + ( x 1 - 2) 1 / (2 n ) , n N . Try to show this and then get the rest of the problem based on this observation (look also at the Example 2.21, page 45 textbook).

This preview has intentionally blurred sections. Sign up to view the full version.

6. (20 pts) Show that a sequence { x n } n is not bounded from above if and only if there exists a subsequence { x n k } k such that x n k → ∞ as k → ∞ . (Note that you have to prove both implications.) Solution: First of all, { x n } n is bounded from above, by definition, if and only if there exists a constant M R , such that x n M, n N . Negating this, { x n } n is not bounded from above if and only if M R , n M N , such that x n M > M. (1) In the above the subscript M for n M , just indicates the dependence on M of the rank. Now let us prove the equivalence asked in the statement. ( ) This implication is easy. Assume that a subsequence { x n k } k → ∞ as k → ∞ . By definition, M R , K N , such that if k K , then x n k > M . Thus relation (1) is trivially satisfied, taking n M to be, for instance n K .
This is the end of the preview. Sign up to access the rest of the document.
• Fall '12
• Draghic
• Mathematical analysis, Order theory, Xn, 1 2k, Dominated convergence theorem, a2 b2

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern